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GATE ECE · Control Systems

Frequency Response Analysis Practice Questions

Generate GATE-level questions on Frequency Response. Focus on: 1. Bode plots: Magnitude and phase plots, determining transfer functions from plots. 2. Nyquist stability criterion: Stability analysis and encirclements. 3. Relative stability: Gain Margin (GM) and Phase Margin (PM). 4. Polar plots and M & N circles.

49 questions · 20 PYQs · 0 AI practice · GATE ECE 2027

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Q1GATE 2025MCQ

The Nyquist plot of a system is given in the figure. Let $\omega_P, \omega_Q, \omega_R$ and $\omega_S$ be the positive frequencies at the points $P, Q, R,$ and $S,$ respectively. Which one of the following statements is TRUE?

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📖 Explanation

From the Nyquist plot: The gain crossover frequency is the frequency at which the Nyquist plot crosses the unit circle, i.e., the magnitude of the open-loop transfer function $|G(j\omega)H(j\omega)| = 1$ . The phase crossover frequency is the frequency at which the phase is $-180^\circ$ , i.e., the Nyquist plot crosses the negative real axis. From the diagram: Point S lies on the unit circle, so $\omega_S$ is the gain crossover frequency. Point P lies on the negative real axis, hence $\omega_P$ is the phase crossover frequency. Therefore, the correct statement is: $\omega_S$ is the gain crossover frequency and $\omega_P$ is the phase crossover frequency.

Q2GATE 2024MCQ

In the context of Bode magnitude plots, $40 \mathrm{~dB} /$ decade is the same as _______

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📖 Explanation

\begin{aligned} 20 \times \mathrm{ndB} / \text { decade } & =6 \times \mathrm{n} \mathrm{dB} / \text { octave } \\ 40 \mathrm{~dB} / \text { decade } & =12 \mathrm{~dB} / \text { octave } \end{aligned}

Q3GATE 2023NAT

The asymptotic magnitude Bode plot of a minimum phase system is shown in the figure.The transfer function of the system is $(s)=\frac{k(s+z)^{a}}{s^{b}(s+p)^{c}}$ , where $k, z, p, a, b$ and $c$ are positive constants. The value of $(a+b+c)$ is ___ (rounded off to the nearest integer).

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📖 Explanation

From the Bode magnitude plot, it is clear that there is one pole at origin, $\therefore$ $b=1$ and at frequency $\omega_{1}$ , system has a zero $\therefore$ $a=1$ and at frequency $\omega_{2}$ , system has a two poles

\begin{aligned} \therefore \quad c&=2 \\ \therefore \quad a+b+c&=1+1+2 \\ a+b+c&=4 \end{aligned}

Q4GATE 2022MCQ

Consider a closed-loop control system with unity negative feedback and $KG(s)$ in the forward path, where the gain $K=2$ . The complete Nyquist plot of the transfer function $G(s)$ is shown in the figure. Note that the Nyquist contour has been chosen to have the clockwise sense. Assume $G(s)$ has no poles on the closed right-half of the complex plane. The number of poles of the closed-loop transfer function in the closed right-half of the complex plane is _______

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📖 Explanation

For K = 2, point A will be -0.8x2 = -1.6 Hence N = -2, P = 0 (By default Nyquist contoure is considered in clockwise direction) P - N = 2 Number of closed loop pole in right side of the complex plane.

Explanation Figure 2

Q5GATE 2021MCQ

The complete Nyquist plot of the open-loop transfer function G(s)H(s) of a feedback control system in the figure. If G(s)H(s) has one zero in the right-half of the s-plane, the number of poles that the closed-loop system will have in the right-half of the s-plane is

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📖 Explanation

The given Nyquist plot is not matched according to the data given in the question.

Q6GATE 2020MCQ

The pole-zero map of a rational function G(s) is shown below. When the closed counter $\Gamma$ is mapped into the G(s)-plane, then the mapping encircles.

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📖 Explanation

s-plane contour is encircling 2-poles and 3-zeros in clockwise direction hence the corresponding G(s) plane contour encircles origin 2-times in anti-clockwise direction and 3-times in clockwise direction. Therefore, Effectively once in clockwise direction.

Q7GATE 2019MCQ

For an LTI system, the Bode plot for its gain is as illustrated in the figure shown. The number of system poles $N_p$ and the number of system zeros $N_z$ in the frequency range $1Hz\leq f\leq 10^7Hz$ is

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📖 Explanation

Number of poles $(N_{P})$ = 6 Number of zeros $(N_{Z})$ = 3

Explanation Figure 2

Q8GATE 2018MCQ

The Nyquist stability criterion and the Routh criterion both are powerful analysis tools for determining the stability of feedback controllers. Identify which of the following statements is FALSE:

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📖 Explanation

Currently no explanation available

Q9GATE 2018NAT

For a unity feedback control system with the forward path transfer function $G(s)=\frac{K}{s(s+2)}$ The peak resonant magnitude $M_{r}$ , of the closed-loop frequency response is 2. The corresponding value of the gain K (correct to two decimal places) is _________.

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📖 Explanation

Maximum resonant peak,

\begin{aligned} M_{r} &=\frac{1}{2 \xi \sqrt{1-\xi^{2}}}=2 \\ 2 \xi \sqrt{1-\xi^{2}} &=\frac{1}{2} \\ \xi^{2}\left(1-\xi^{2}\right) &=\frac{1}{16} \\ \xi^{4}-\xi^{2}+\frac{1}{16} &=0 \\ \xi^{2}&=\frac{1}{2} \pm \sqrt{\frac{1-\frac{1}{4}}{4}}=\frac{1}{2} \pm \frac{\sqrt{3}}{4} \\ \text { As } M_{r}=2 \gt 1, \xi& \lt \frac{1}{\sqrt{2}} \text { and } \xi^{2} \lt \frac{1}{2} \\ \text { So. } \quad\xi^{2}&=\frac{1}{2}-\frac{\sqrt{3}}{4}\\ Given, \quad G(s)&=\frac{K}{s(s+2)}=\frac{\omega_{n}^{2}}{s\left(s+2 \xi \omega_{n}\right)} \\ \text{So} \quad\omega_{n} &=\sqrt{K} \\ 2 \xi \sqrt{K} &=2\\ \sqrt{K} &=\frac{1}{\xi} \\ K &=\frac{1}{\xi^{2}}=\frac{1}{\left(\frac{1}{2}-\frac{\sqrt{3}}{4}\right)} \\ &=\frac{4}{2-\sqrt{3}}=14.928 \end{aligned}

Q10GATE 2018NAT

The figure below shows the Bode magnitude and phase plots of a stable transfer function $G(s)=\frac{n_{o}}{s^{2}+d_{2}s^{2}+d_{1}+d_{0}}$ Consider the negative unity feedback configuration with gain k in the feedforward path. The closed loop is stable for $k \lt k_{o}$ . The maximum value of $k_{o}$ is ______.

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📖 Explanation

For G(s) $M_{\mathrm{dB}}\left(\omega_{p c}\right)=20 \mathrm{dB}$ When cascaded with k ,

\begin{aligned} \mathrm{GM}_{\mathrm{dB}} &=-20 \mathrm{dB}-20 \log _{10}(k) \gt 0 \mathrm{dB} \\ 20+20 \log _{10}(k) & \lt 0 \\ 20 \log _{10}(k) & \lt -20 \\ k & \lt 10^{-1}=0.10\\ \text{So,}\quad k_{0}&=0.10 \end{aligned}

Q11GATE 2017MCQ

A unity feedback control system is characterized by the open-loop transfer function $G(s)=\frac{10K(s+2)}{s^{3}+3s^{2}+10}$ The Nyquist path and the corresponding Nyquist plot of G(s) are shown in the figures below. If $0 \lt K \lt 1$ , then the number of poles of the closed-loop transfer function that lie in the right half of the s-plane is

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📖 Explanation

Given that, the open loop transfer function of a unity feedback system is, $G(s) =\frac{10 k(s+2)}{s^{3}+3 s^{2}+10}$ From the given Nyquist plot, for $0 \lt k \lt 1$ the encirclements about the point (-1+j 0) is,

\begin{array}{l} N=0 \\ N=P-Z \end{array}

P= number of open loop poles in the right half of s-plane Z= number of closed loop poles in the right half of s-plane To determine the value of P : Applying R-H criteria to G(s) $G(s)=\frac{10 k(s+2)}{s^{3}+3 s^{2}+10}$

\begin{array}{c|cc} s^{3} & 1 & 0 \\ s^{2} & 3 & 10 \\ s^{1} & -10 / 3 & 0 \\ s^{0} & 10 & 0 \end{array}

Two sign changes are there in the first column of the RH table. So, two open loop poles are there in right half of s-plane. So, $P=2$

\begin{array}{l} N=P-Z \\ 0=2-Z \\ Z=2 \end{array}

Z=2 indicates, there are two closed loop poles in right half of s-plane.

Q12GATE 2017MCQ

Consider a stable system with transfer function $G(s)=\frac{s^{p}+b_{1}S^{p-1}+....+b_p}{s^{q}+a_{1}S^{q-1}+....+a_{q}}$ where $b_{1},...b_{p}$ and $a_{1},...a_{q}$ are real valued constants. The slope of the Bode log magnitude curve of G(s) converges to -60 dB/decade as $\omega \rightarrow \infty$ . A possible pair of values for p and q is

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📖 Explanation

Final slope = -60 dB/decade, which indicates that, q-p=3 Among the given options, option (A) satisfies this condition

Q13GATE 2017MCQ

The Nyquist plot of the transfer function $G(s)=\frac{K}{(s^{2}+2s+2)(s+2)}$ does not encircle the point (1+j0) for K=10 but does encircle the point (-1+j0) for K=100. Then the closed loop system (having unity gain feedback) is

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📖 Explanation

Given that, $G(s)=\frac{K}{\left(s^{2}+2 s+2\right)(s+2)}$ P= open loop poles in RHS of s-plane Z= closed loop poles in RHS of s-plane N= number of encirclements about the

\begin{array}{l} \text { point }(-1+j 0) \\ \text { For } K=10 ; \\ N=P-Z=0 \Rightarrow Z=0: \text { stable } \\ \text { For } K=100 ; \\ N=P-Z=1 \Rightarrow Z \neq 0: \text { unstable } \end{array}

Q14GATE 2016MCQ

A closed-loop control system is stable if the Nyquist plot of the corresponding open-loop transfer function

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📖 Explanation

N= P-Z N = Number of encirclements of (-1 + /J). It is positive if nyquist plot encircles the point -1 + j0 in counterclockwise direction. Z = Number of closed loop poles lying in the right half of s-plane P = Number of open loop poles lying in right half of s-plane For stability Z = 0 $\Rightarrow$ N = P

Q15GATE 2016NAT

The asymptotic Bode phase plot of $G(s)=\frac{1}{(s+0.1)(s+10)(s+p_{1})}$ , with k and $p_{1}$ both positive, is shown below. The value of $p_{1}$ is ________

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📖 Explanation

\begin{aligned} G(s)=\frac{K}{(s+0.1)(s+10)\left(s+p_{1}\right)} \\ \angle G(s)=-\tan ^{-1} \frac{\omega}{0.1}-\tan ^{-1} \frac{\omega}{10}-\tan ^{-1} \frac{\omega}{p_{1}} \\ \left.\angle G(s)\right|_{\omega=1}=-\tan ^{-1} \frac{1}{0.1}-\tan ^{-1} \frac{1}{10}-\tan ^{-1} \frac{1}{p_{1}} \\ =-135^{\circ} \\ -\tan ^{-1} 10-\tan ^{-1} 0.1-\tan ^{-1} \frac{1}{p_{1}}=-135^{\circ} \\ -84.28^{\circ}-5.71^{\circ}-\tan ^{-1} \frac{1}{p_{1}}=-135^{\circ} \\ -\tan ^{-1} \frac{1}{p_{1}}=-135^{\circ}+90^{\circ}\\ \tan ^{-1} \frac{1}{p_{1}} =45^{\circ} \\ \frac{1}{p_{1}} =1 \Rightarrow p_{1}=1 \end{aligned}

Q16GATE 2016NAT

In the feedback system shown below $G(s)=\frac{1}{(s+1)(s+2)(s+3)}$ The positive value of k for which the gain margin of the loop is exactly 0 dB and the phase margin of the loop is exactly zero degree is ________

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📖 Explanation

$1+G(s) H(s)=1+\frac{K}{(s+1)(s+2)(s+3)}=0$ $(s+1)\left(s^{2}+5 s+6\right)+K=0$ $s^{3}+5 s^{2}+6 s+s^{2}+5 s+6+k=0$ $\Rightarrow s^{3}+6 s^{2}+11 s+6+K=$ Gain margin =0dB and phase margin $=0^{\circ}$ It implies marginal stable system By Routh Array

\begin{array}{r|rr} s^{3} & 1 & 11 \\ s^{2} & 6 &(6+k) \\ s & \frac{66-6-K}{6} & 0 \\ s^{\circ} & 6+K & \end{array}

For marginal stable system, $60-K=0$ $K=60$

Q17GATE 2016MCQ

The number and direction of encirclements around the point -1+j0 in the complex plane by the Nyquist plot of $G(s)=\frac{1-s}{4+2s}$ is

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📖 Explanation

$G(s)=\frac{1-s}{4+2 s}$ and $\angle G(s)=-\tan ^{-1} \omega-\tan ^{-1} \frac{\omega}{2}$ at $s=0, \quad|G(s)|=\frac{1}{4}=0.25$ and $\left.\angle G(s)\right|_{\omega=0}=0^{\circ}$ at $s=\infty , \quad|G(s)|=-\frac{1}{2}=-0.5$ and $\left.\angle G(s)\right|_{m \times \infty }=-180^{\circ}$ Nyquist plot is Hence, number of encirclements of (-1 + j0) = 0

Explanation Figure 1

Q18GATE 2015NAT

The phase margin (in degrees) of the system $G(s)=\frac{10}{s(s+10)}$ is _______.

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📖 Explanation

\begin{aligned} G(s) &=\frac{10}{s(s+10)} \\ P M &=180^{\circ}+\left.\angle G(\omega) H(\omega)\right|_{\omega=\omega_{g c}} \end{aligned}

The gain cross over frequency can be calculated by

\begin{aligned} |G(\omega) H(\omega)|&=1 \\ \frac{10}{\omega \sqrt{\omega^{2}+100}}&=1 \\ \omega_{g c}^{4}+100 \omega_{g c}^{2}&-100=0\\ \text { Or } \quad \omega_{g c}^{2}&=0.99, \quad-100.99 \\ \text { Or } \quad \omega_{g c}&=0.99 \text { rad/sec. } \\ \therefore P M &=180^{\circ}+\left(-90^{\circ}-\tan ^{-1} \frac{\omega_{g c}}{10}\right) \\ &=180^{\circ}-90^{\circ}-\tan ^{-1} \frac{0.99}{10} \\ & \simeq 84.364^{\circ} \end{aligned}

Q19GATE 2015NAT

The transfer function of a mass-spring-damper system is given by $G(s)=\frac{1}{Ms^{2}+Bs+K}$ The frequency response data for the system are given in the following table. The unit step response of the system approaches a steady state value of ________ .

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📖 Explanation

$\frac{\alpha(s)}{R(s)}=G(s)=\frac{1}{M s^{2}+B s+K}$ Steady state value of response for unit step input

\begin{aligned} C(s)&=\lim _{s \rightarrow 0} s \times \frac{1}{M s^{2}+B s+K} \times \frac{1}{s}=\frac{1}{K} \\ \text{Also}&\\ G(\approx 0)&=\frac{1}{K}=-18.5 \mathrm{dB} \\ &\qquad\text{(as per the given table)}\\ \therefore \frac{1}{K}&=10^{\left(-\frac{18.5}{20}\right)}=0.12 \end{aligned}

Q20GATE 2015NAT

Consider the Bode plot shown in the figure. Assume that all the poles and zeros are real-valued. The value of $f_{H}-f_{L}$ (in Hz) is ___________.

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📖 Explanation

As we know,

\begin{aligned} \text { Slope } &=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\ \therefore \quad 40 \mathrm{dB} / \mathrm{dec} &=\frac{40-0}{\log 300-\log _{L}}\\ or, \log \left(\frac{300}{f_{L}}\right)&=\frac{40}{40}=1 f_{L}=30 \mathrm{Hz} \end{aligned}

Similarly,

\begin{aligned} -40 \mathrm{dB} / \mathrm{dec}&=\frac{40-0}{\log 900-\log f_{H}}\\ \log \left(\frac{900}{f_{H}}\right) &=-1 \\ \text { or, } \quad f_{H}&=9000 \mathrm{Hz} \\ \text { Therefore, } f_{H}-f_{L}&=9000-30 =8970 \mathrm{Hz} \end{aligned}

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